Answer to Question #197699 in Electric Circuits for Plea

Question

A proton moves from point A to point B. Its distance from point A is 2mm and from point B is 4 mm. Find: a. the potential difference from point A to B b. work done on the proton

Answer 1

Gives

Point P is another point

AP=2mm

BP=4mm

q=$1.6\times10^{-19}c$

Particle is Proton

Due to poin A potential difference

$V_A=\frac{kq}{r^2}$

$V_A=\frac{9\times10^{9}\times1.6\times10^{-19}}{2\times10^{-3}}$

$V_A=7.2\times10^{-7}V$

$V_B=\frac{kq}{r^2}$

$V_B=\frac{9\times10^9\times1.6\times10^{-19}}{4\times10^{-3}}$

$V_B=3.6\times10^{-7}V$

Potential difference between A to B$V_{AB}=V_A-V_B$

$V_{AB}={7.2 \times10^{-7}}-{3.6\times10^{-7}}$

$V_{AB}=3.6\times10^{-7}V$

Part (b)

Work done by proton

$W=q∆V$

$W=1.6\times10^{-19}\times3.6\times10^{-7}$

$W=5.76\times10^{-26}J$